What are the steps to solve (sec∂-tan∂)²=(1-sin∂)/(1+sin∂)?

For 2. Sec=1/cos tan=sin/cos cos²x+sin²x=1:Use the Pythagorean identity, which states that cos²(x) + sin²(x) = 1. Therefore, cos²(x) + sin²(x) = 1 is always true.
  • #1
chris99191
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1. First one is (sin2x+sinx)/(cos2x+cosx+1)=tanx
Second one is (sec∂-tan∂)²=(1-sin∂)/(1+sin∂)


2. Sec=1/cos tan=sin/cos cos²x+sin²x=1



3. 1. I think eventually the sinx/cosx need to cancel to make tanx and the 1 could be used to create a lot of options
2. I have tried to start with the left hand side. However i am unsure whether to expand the bracket or not. I've tried it with 1/cos but I couldn't get very far
 
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  • #2
chris99191 said:
1. ...
Second one is (sec∂-tan∂)²=(1-sin∂)/(1+sin∂)

3. 1. I think eventually the sinx/cosx need to cancel to make tanx and the 1 could be used to create a lot of options
2. I have tried to start with the left hand side. However i am unsure whether to expand the bracket or not. I've tried it with 1/cos but I couldn't get very far


For 1. Second one:

Use sec(x) = 1/cos(x) and tan(x) = sin(x)/cos(x). They have a common denominator, so combine them into one fraction. After squaring, change cos2(∂) to 1 - sin2(∂) .
 

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